3.2.50 \(\int x^m (d-c^2 d x^2)^{3/2} (a+b \arcsin (c x)) \, dx\) [150]

3.2.50.1 Optimal result

Integrand size = 27, antiderivative size = 399 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1-c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d-c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{4+m} \sqrt {d-c^2 d x^2}}{(4+m)^2 \sqrt {1-c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{4+m}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1-c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d-c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) \sqrt {1-c^2 x^2}} \]

output

x^(1+m)*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/(4+m)+3*d*x^(1+m)*(a+b*arcs 
in(c*x))*(-c^2*d*x^2+d)^(1/2)/(m^2+6*m+8)-3*b*c*d*x^(2+m)*(-c^2*d*x^2+d)^( 
1/2)/(2+m)^2/(4+m)/(-c^2*x^2+1)^(1/2)-b*c*d*x^(2+m)*(-c^2*d*x^2+d)^(1/2)/( 
m^2+6*m+8)/(-c^2*x^2+1)^(1/2)+b*c^3*d*x^(4+m)*(-c^2*d*x^2+d)^(1/2)/(4+m)^2 
/(-c^2*x^2+1)^(1/2)+3*d*x^(1+m)*(a+b*arcsin(c*x))*hypergeom([1/2, 1/2+1/2* 
m],[3/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/(m^3+7*m^2+14*m+8)/(-c^2*x^2+ 
1)^(1/2)-3*b*c*d*x^(2+m)*hypergeom([1, 1+1/2*m, 1+1/2*m],[2+1/2*m, 3/2+1/2 
*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/(2+m)^2/(m^2+5*m+4)/(-c^2*x^2+1)^(1/2)
 

3.2.50.2 Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.59 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\frac {d x^{1+m} \sqrt {d-c^2 d x^2} \left (-\frac {b c x \left (4+m-c^2 (2+m) x^2\right )}{(2+m) (4+m) \sqrt {1-c^2 x^2}}+\left (1-c^2 x^2\right ) (a+b \arcsin (c x))-\frac {3 \left (b c (1+m) x-(1+m) (2+m) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-(2+m) (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )+b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )\right )}{(1+m) (2+m)^2 \sqrt {1-c^2 x^2}}\right )}{4+m} \]

input

Integrate[x^m*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]
 

output

(d*x^(1 + m)*Sqrt[d - c^2*d*x^2]*(-((b*c*x*(4 + m - c^2*(2 + m)*x^2))/((2 
+ m)*(4 + m)*Sqrt[1 - c^2*x^2])) + (1 - c^2*x^2)*(a + b*ArcSin[c*x]) - (3* 
(b*c*(1 + m)*x - (1 + m)*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) - ( 
2 + m)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^ 
2*x^2] + b*c*x*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/ 
2}, c^2*x^2]))/((1 + m)*(2 + m)^2*Sqrt[1 - c^2*x^2])))/(4 + m)
 

3.2.50.3 Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5202, 244, 2009, 5198, 15, 5220}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx\)

\(\Big \downarrow \) 5202

\(\displaystyle \frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))dx}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \int x^{m+1} \left (1-c^2 x^2\right )dx}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}\)

\(\Big \downarrow \) 244

\(\displaystyle \frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))dx}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \int \left (x^{m+1}-c^2 x^{m+3}\right )dx}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))dx}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2}}{m+2}-\frac {c^2 x^{m+4}}{m+4}\right )}{(m+4) \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 5198

\(\displaystyle \frac {3 d \left (\frac {\sqrt {d-c^2 d x^2} \int \frac {x^m (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{(m+2) \sqrt {1-c^2 x^2}}-\frac {b c \sqrt {d-c^2 d x^2} \int x^{m+1}dx}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{m+2}\right )}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2}}{m+2}-\frac {c^2 x^{m+4}}{m+4}\right )}{(m+4) \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {3 d \left (\frac {\sqrt {d-c^2 d x^2} \int \frac {x^m (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 \sqrt {1-c^2 x^2}}\right )}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2}}{m+2}-\frac {c^2 x^{m+4}}{m+4}\right )}{(m+4) \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 5220

\(\displaystyle \frac {3 d \left (\frac {\sqrt {d-c^2 d x^2} \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{m+1}-\frac {b c x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{m^2+3 m+2}\right )}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+2} \sqrt {d-c^2 d x^2}}{(m+2)^2 \sqrt {1-c^2 x^2}}\right )}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{m+4}-\frac {b c d \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2}}{m+2}-\frac {c^2 x^{m+4}}{m+4}\right )}{(m+4) \sqrt {1-c^2 x^2}}\)

input

Int[x^m*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]
 

output

-((b*c*d*Sqrt[d - c^2*d*x^2]*(x^(2 + m)/(2 + m) - (c^2*x^(4 + m))/(4 + m)) 
)/((4 + m)*Sqrt[1 - c^2*x^2])) + (x^(1 + m)*(d - c^2*d*x^2)^(3/2)*(a + b*A 
rcSin[c*x]))/(4 + m) + (3*d*(-((b*c*x^(2 + m)*Sqrt[d - c^2*d*x^2])/((2 + m 
)^2*Sqrt[1 - c^2*x^2])) + (x^(1 + m)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x 
]))/(2 + m) + (Sqrt[d - c^2*d*x^2]*((x^(1 + m)*(a + b*ArcSin[c*x])*Hyperge 
ometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(1 + m) - (b*c*x^(2 + m)*H 
ypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(2 
 + 3*m + m^2)))/((2 + m)*Sqrt[1 - c^2*x^2])))/(4 + m)
 

3.2.50.3.1 Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5198

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + 
(e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcS 
in[c*x])^n/(f*(m + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt[1 
 - c^2*x^2]]   Int[(f*x)^m*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x 
] - Simp[b*c*(n/(f*(m + 2)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[ 
(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, 
 f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])
 

rule 5202

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcS 
in[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1))   Int[(f*x) 
^m*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Simp[b*c*(n/(f*(m + 
2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 - c^2 
*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, 
e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
 

rule 5220

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_. 
)*(x_)^2], x_Symbol] :> Simp[((f*x)^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2* 
x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, 
 (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*S 
imp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m 
/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e, f, m}, x] 
&& EqQ[c^2*d + e, 0] &&  !IntegerQ[m]
 

3.2.50.4 Maple [F]

\[\int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )d x\]

input

int(x^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x)
 

output

int(x^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x)
 

3.2.50.5 Fricas [F]

\[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]

input

integrate(x^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="fricas" 
)
 

output

integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))*sqrt(-c^2* 
d*x^2 + d)*x^m, x)
 

3.2.50.6 Sympy [F(-1)]

Timed out. \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]

input

integrate(x**m*(-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x)),x)
 

output

Timed out
 

3.2.50.7 Maxima [F]

\[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]

input

integrate(x^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="maxima" 
)
 

output

integrate((-c^2*d*x^2 + d)^(3/2)*(b*arcsin(c*x) + a)*x^m, x)
 

3.2.50.8 Giac [F(-2)]

Exception generated. \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\text {Exception raised: TypeError} \]

input

integrate(x^m*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

3.2.50.9 Mupad [F(-1)]

Timed out. \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]

input

int(x^m*(a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2),x)
 

output

int(x^m*(a + b*asin(c*x))*(d - c^2*d*x^2)^(3/2), x)